Calculating K_a from ΔE° for HBrO Reaction

[amcavoy]

Using $$\Delta \textrm{E}^{\circ}$$ values and the fact that $$\textrm{K}_{\textrm{w}}=10^{-14}$$, how would I find the $$\textrm{K}_{\textrm{a}}$$ value for the following reaction?:

$$\textrm{HBrO}\longrightarrow\textrm{H}^{+}+\textrm{BrO}^{-}$$​

 

[Noober]

dEº = -RTlnK

Dunno if that helps :/.

 

[Weave]

Noobler sounds right.
Rearrange to give:
$$\frac{\Delta E}{-RT}=ln K_{a}$$
and solve for $$K_{a]$$